Question

当我擦除向量中的元素时，我收到奇怪的分段错误错误。这是代码：

int i = 0;
    cout << "SIZE: " << OrderList::sellOrders.size();
    for(vector<Order>::iterator it = OrderList::sellOrders.begin(); it != OrderList::sellOrders.end(); it++, i++)
    {
        if(OrderList::sellOrders[i].pThis->usesWamp() == false)
        {
            cout << "Erase element " << i << endl;
            OrderList::sellOrders.erase(it);
        }
    }

输出是这样的：

SIZE: 44
Erase element 0
Erase element 2
Erase element 3
Erase element 4
Erase element 5
Erase element 6
Erase element 7
Erase element 8
Erase element 9
Erase element 10
Erase element 11
Erase element 12
Erase element 13
Erase element 14
Erase element 15
Erase element 16
Erase element 17
Erase element 18
Erase element 19
Erase element 20
Erase element 21
Erase element 22
Erase element 23

Program received signal SIGSEGV, Segmentation fault.
__memmove_ssse3_back () at ../sysdeps/x86_64/multiarch/memcpy-ssse3-back.S:1548
1548    ../sysdeps/x86_64/multiarch/memcpy-ssse3-back.S: No such file or directory.
(gdb)
(gdb) where
#0  __memmove_ssse3_back ()
    at ../sysdeps/x86_64/multiarch/memcpy-ssse3-back.S:1548
#1  0x0000000000485d48 in std::__copy_move<true, true, std::random_access_iterator_tag>::__copy_m<Order> (__first=0x7fffe400e4a0, __last=0x7fffe400e470,
    __result=0x7fffe400e488) at /usr/include/c++/4.8/bits/stl_algobase.h:372
#2  0x00000000004807c4 in std::__copy_move_a<true, Order*, Order*> (
    __first=0x7fffe400e4a0, __last=0x7fffe400e470, __result=0x7fffe400e488)
    at /usr/include/c++/4.8/bits/stl_algobase.h:390
#3  0x000000000047879f in std::__copy_move_a2<true, __gnu_cxx::__normal_iterator<Order*, std::vector<Order, std::allocator<Order> > >, __gnu_cxx::__normal_iterator<Order*, std::vector<Order, std::allocator<Order> > > > (__first=...,
    __last=..., __result=...) at /usr/include/c++/4.8/bits/stl_algobase.h:428
#4  0x000000000046f9d3 in std::move<__gnu_cxx::__normal_iterator<Order*, std::vector<Order, std::allocator<Order> > >, __gnu_cxx::__normal_iterator<Order*, std::vector<Order, std::allocator<Order> > > > (__first=..., __last=...,
    __result=...) at /usr/include/c++/4.8/bits/stl_algobase.h:492
#5  0x0000000000467754 in std::vector<Order, std::allocator<Order> >::erase (
    this=0x6c7c10 <OrderList::sellOrders>, __position=...)
    at /usr/include/c++/4.8/bits/vector.tcc:138
#6  0x0000000000421320 in OrderList::flush () at ../Bether/OrderList.h:339
#7  0x000000000044b12a in main () at ../Bether/main.cpp:354

因此，您可以看到Vector有44个元素（SIZE: 44），但在删除元素23时失败。

Answer 1

std::vector::erase()使迭代器无效，这意味着以后取消引用它（*it）或增加它（++it）是未定义的行为。

一个很好的和通常的范例是合并std::remove_if()移动容器背面的移除项目std::vector::erase()实际删除它们：

std::vector<int> v = { 0, 1, 2, 3, 4, 5, 6 };
v.erase( std::remove_if(v.begin(), v.end(), [](int n){ return n%2==0; }), v.end() );
// v is { 1, 3, 5 }

Answer 2

Enumerable.FirstOrDefault函数使指向删除位置的迭代器和引用无效。

如果你真的想要删除它们，你可以改变循环以按递减顺序迭代，这应该没有问题，因为你使用的向量很容易允许双向访问。这样，只有对已检查过的位置的引用才会失效。

Vector.erase中的SIGSEGV

2 个答案: